A proton is bombarded on a stationary lithium nucleus. As a result...

Q value of the reaction is,

Q = (2 × 4 × 7.06 – 7 × 5.6) MeV = 17.28 MeV



Applying conservation of energy for collision,

K_p + Q = 2 K_α ....(i)

(Here, K_p and K_α are the kinetic energies of proton and α - particle respectively)

From the conservation of linear momentum (As there is no external force) ....(ii)

$[ Here P = \sqrt{2mk} ]$
${⇒ K }_{p } = 1 6 {K }_{\alpha} {\text{cos}}^{2} \theta = \left(1 6 {K }_{\alpha}\right) {\left(\dfrac{1}{4}\right)}^{2} \left(\text{as} {m }_{\alpha} = 4 {m }_{p }\right)$

∴ K_α =K_{p ....(iii)}

Solving eqs. (i) and (iii) with Q = 17.28 MeV

we get K_p = 17.28 MeV